Integral Calculus: Basic Integration of cotxln(sinx)

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In summary, using basic integration techniques from calculus 1, the integral \displaystyle \int \cot(x)\ln(\sin(x))\,dx can be evaluated by substituting u=sinx and using the formula \int \frac{\ln(u)}{u} du = \frac{[ln(u)]^2}{2} + C, resulting in the final answer of \frac{[ln(sinx)]^2}{2} + C.
  • #1
whatlifeforme
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Homework Statement


evaluate the integral.

Homework Equations


[itex]\int (cotx)[ln(sinx)] dx[/itex]

how would i do this using basic integration techniques from calculus 1? i am not allowed to use, int by parts, partial fractions, trig sub, etc..

The Attempt at a Solution


i tried doing a u=sinx but don't know how to get the cotx.
 
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  • #2
whatlifeforme said:

Homework Statement


evaluate the integral.

Homework Equations


[itex]\int (cotx)[ln(sinx)][/itex]

how would i do this using basic integration techniques from calculus 1? i am not allowed to use, int by parts, partial fractions, trig sub, etc..

The Attempt at a Solution


i tried doing a u=sinx but don't know how to get the cotx.
Where's the "dx" ?

[itex]\displaystyle \int \cot(x)\ln(\sin(x))\,dx[/itex]

This may help. [itex]\displaystyle \ \cot(x)=\frac{\cos(x)}{\sin(x)}[/itex]
 
  • #3
thanks.

so i have:

u=sinx; du=cosx

[itex]\int \frac{ln(u)}{u} du[/itex]

v=lnu; dv=(1/u) du

[itex]\int v dv[/itex] = [itex]\frac{v^2}{2}[/itex] = [itex]\frac{(lnu)^2}{2}[/itex]

= [itex]\frac{[ln(sinx)]^2}{2}[/itex]
 
  • #4
whatlifeforme said:
thanks.

so i have:

u=sinx; du=cosx

[itex]\int \frac{ln(u)}{u} du[/itex]

v=lnu; dv=(1/u) du

[itex]\int v dv[/itex] = [itex]\frac{v^2}{2}[/itex] = [itex]\frac{(lnu)^2}{2}[/itex]

= [itex]\frac{[ln(sinx)]^2}{2}[/itex]

Looks good to me, other than the +C.
 

Related to Integral Calculus: Basic Integration of cotxln(sinx)

1. What is the purpose of using cotxln(sinx) in integral calculus?

The function cotxln(sinx) is commonly used in integral calculus to evaluate the area under a curve. It is also a useful tool for finding the antiderivative of certain functions.

2. What is the general process for integrating cotxln(sinx)?

The general process for integrating cotxln(sinx) involves using the substitution method. This means replacing cotx with its reciprocal, tanx, and ln(sinx) with its equivalent form of -ln(cosx). The resulting integral can then be solved using standard integration techniques.

3. Are there any special cases when integrating cotxln(sinx)?

Yes, there are a few special cases to consider when integrating cotxln(sinx). If the limits of integration are from 0 to π/2, then the integral will be equal to 0. Additionally, if the limits are from π/2 to π, then the integral will be equal to -πln(2).

4. Can cotxln(sinx) be integrated using other methods besides substitution?

Yes, there are other methods that can be used to integrate cotxln(sinx), such as integration by parts or trigonometric identities. However, the substitution method is often the most efficient and straightforward approach.

5. How can understanding the integration of cotxln(sinx) be applied to real-world problems?

The integration of cotxln(sinx) can be applied to a variety of real-world problems, such as calculating the work done by a varying force or finding the arc length of a cycloid curve. It is also a fundamental tool in engineering, physics, and other scientific fields.

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